Question

The line joining the points $6\overrightarrow{a}-4\overrightarrow{b}-5\overrightarrow{c},-4\overrightarrow{c}$ and the line joining the points $-\overrightarrow{a}-2\overrightarrow{b}-3\overrightarrow{c},\overrightarrow{a}+2\overrightarrow{b}-5\overrightarrow{c}$ intersect at

• A. $2\overrightarrow{c}$
• B. $-4\overrightarrow{c}$
• C. $8\overrightarrow{c}$
• D. None of these

Answer 1

Answer: (B)

$-4\overrightarrow{c}$

Equation of the line joining $6\overrightarrow{a}-4\overrightarrow{b}-5\overrightarrow{c}$ and $-4\overrightarrow{c}$ is

$\overrightarrow{r}=(6\overrightarrow{a}-4\overrightarrow{b}-5\overrightarrow{c})+\lambda (-6\overrightarrow{a}+4\overrightarrow{b}+\overrightarrow{c})$

$\overrightarrow{r}=\overrightarrow{a}(6-6\lambda )+\overrightarrow{b}(-4+4\lambda )+\overrightarrow{c}(-5+\lambda )$ .....(1)

Also equation of the line joining $-\overrightarrow{a}-2\overrightarrow{b}-3\overrightarrow{c}$ and $\overrightarrow{a}+2\overrightarrow{b}-5\overrightarrow{c}$

$\overrightarrow{r}=(-\overrightarrow{a}-2\overrightarrow{b}-3\overrightarrow{c})+\mu (2\overrightarrow{a}+4\overrightarrow{b}-2\overrightarrow{c})$

$\Rightarrow \overrightarrow{r}=\overrightarrow{a}(-1+2\mu )+\overrightarrow{b}(-2+4\mu )+\overrightarrow{c}(-3-2\mu )$ ....(2)

Compare $1$ and $2$ to get

$\Rightarrow 6-6\lambda =(-1+2\mu ),(-2+4\mu )=-4+4\lambda ,(-3-2\mu )=-5+\lambda$

Solve these equations to get $\lambda =1,\mu =\frac{1}{2}$

Substitute value of $\lambda =1$ in $1$ to get point of intersection as $\overrightarrow{r}=-4\overrightarrow{c}$